Common Ion Effect

AP Chemistry· AP Chemistry CED — Equilibrium· 14 min read

1. Definition and Core Concept★☆☆☆☆⏱ 2 min

The common ion effect describes the suppression of ionization or dissolution of a weak electrolyte (weak acid, weak base, or sparingly soluble ionic salt) when a soluble compound containing an ion already present in the equilibrium (the "common ion") is added to the solution. It is a direct application of Le Chatelier’s principle to equilibrium systems. Unlike minor ionic strength effects from adding non-common ion salts (which are not tested on AP Chemistry), the common ion effect produces a large, easily calculable change in equilibrium concentrations that is explicitly tested on the exam. This topic accounts for 7-9% of the total AP exam score, appears in both MCQ and FRQ, and is the foundation of buffer solutions.

📘 Definition

Common Ion Effect

Suppression of ionization or dissolution of a weak electrolyte caused by adding a soluble compound containing an ion already present in the weak electrolyte's equilibrium.

2. Effect on Weak Acid/Base Ionization★★☆☆☆⏱ 4 min

When a weak acid $H A$ ionizes in solution, it establishes the equilibrium:

HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)

If a soluble salt of the conjugate base $N a A$ is added, it dissociates completely to release $A^{-}$ , the common ion. Adding $A^{-}$ increases product concentration, shifting equilibrium left toward undissociated $H A$ . This reduces percent ionization of $H A$ , lowers $[H^{+}]$ , and increases pH compared to a solution of $H A$ alone. The same logic applies to weak bases: adding a common conjugate acid suppresses ionization, lowers $[O H^{-}]$ , and decreases pH.

We use the small x (5%) approximation, which assumes dissociation of the weak acid is negligible compared to initial concentrations of $H A$ and added $A^{-}$ :

K_a = \frac{[H^+][A^-]}{[HA]} \approx \frac{[H^+][A^-]_0}{[HA]_0}

Rearranging gives the Henderson-Hasselbalch equation, a direct result of the common ion effect:

pH = pK_a + \log\left(\frac{[A^-]_0}{[HA]_0}\right)

📐 Worked Example

Calculate the pH of a solution that is 0.15 M hydrofluoric acid (HF, $K_{a} = 6.6 \times 1 0^{- 4}$ ) and 0.25 M sodium fluoride (NaF).

1
Identify the common ion: $F^{-}$ is common to both HF and NaF. NaF dissociates completely, so initial $[F^{-}]_{0} = 0.25$ M, $[H F]_{0} = 0.15$ M.
2
Let $x = [H^{+}]$ from HF dissociation. Equilibrium concentrations are:
3
$[HF] = 0.15 - x, \quad [H^+] = x, \quad [F^-] = 0.25 + x$
4
Apply the 5% approximation: $x << 0.15$ and $x << 0.25$ , so concentrations simplify. Substitute into $K_{a}$ :
5
$6.6 \times 10^{-4} = \frac{(x)(0.25)}{0.15}$
6
Solve for $x = [H^{+}] = 3.96 \times 1 0^{- 4}$ M. Check approximation: $x = 0.16%$ of 0.25 M, so approximation is valid.
7
Calculate pH:
8
$pH = -\log(3.96 \times 10^{-4}) \approx 3.40$

Exam tip:

On the AP exam, you can directly use the Henderson-Hasselbalch equation for common ion weak acid/base systems to save time, as the 5% approximation almost always holds.

3. Effect on Solubility of Sparingly Soluble Salts★★★☆☆⏱ 5 min

The common ion effect significantly reduces the molar solubility ( $s$ , moles of solid that dissolve per liter of solution) of sparingly soluble ionic compounds. For the general dissolution equilibrium:

M_xX_y(s) \rightleftharpoons xM^{y+}(aq) + yX^{x-}(aq)

If a common ion is added, equilibrium shifts left toward the solid, reducing solubility. For sparingly soluble salts, $s$ is already very small, so the contribution of dissolved salt to the common ion concentration is negligible, so we approximate the common ion concentration as equal to its initial added concentration.

📐 Worked Example

Calculate the molar solubility of calcium oxalate ( $C a C_{2} O_{4}$ , $K_{s p} = 2.3 \times 1 0^{- 9}$ ) in a 0.15 M solution of calcium chloride ( $C a C l_{2}$ ).

1
Write the balanced dissolution equilibrium. The common ion is $C a^{2 +}$ from complete dissociation of $C a C l_{2}$ :
2
$CaC_2O_4(s) \rightleftharpoons Ca^{2+}(aq) + C_2O_4^{2-}(aq)$
3
Let $s$ = molar solubility of $C a C_{2} O_{4}$ . Equilibrium concentrations are:
4
$[Ca^{2+}] = 0.15 + s, \quad [C_2O_4^{2-}] = s$
5
Apply the small $s$ approximation: $s << 0.15$ , so $0.15 + s \approx 0.15$ . Substitute into $K_{s p}$ :
6
$2.3 \times 10^{-9} = (0.15)(s)$
7
Solve for $s$ : $s \approx 1.5 \times 1 0^{- 8}$ M. The approximation is valid, as $s$ is orders of magnitude smaller than 0.15 M. For comparison, solubility in pure water is ~ $4.8 \times 1 0^{- 5}$ M, 3000x higher, confirming strong solubility suppression.

Exam tip:

Always write the balanced dissolution reaction before plugging concentrations into $K_{s p}$ ; do not assume all salts are 1:1.

4. Qualitative Predictions★★☆☆☆⏱ 3 min

AP exams commonly ask for qualitative predictions of how adding a common ion changes pH, percent ionization, or solubility, with no calculation required. These rely on core rules:

Adding a common product ion always shifts equilibrium toward the reactant side.
No common ion = no common ion effect (ionic strength effects are not tested).
For weak acids: adding common conjugate base → lower $[H^{+}]$ → higher pH → lower percent ionization.
For weak bases: adding common conjugate acid → lower $[O H^{-}]$ → lower pH → lower percent ionization.
For sparingly soluble salts: adding any common ion → lower solubility.

📐 Worked Example

For each change to a 0.10 M acetic acid solution (initial pH = 2.87), state whether pH increases, decreases, or stays the same, and justify: (a) Solid sodium acetate is added. (b) Solid sodium chloride is added.

1
Write the acetic acid ionization equilibrium:
2
$CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)$
3
For (a): Sodium acetate releases $C H_{3} C O O^{-}$ , the common product ion. Equilibrium shifts left, reducing $[H^{+}]$ , so pH increases.
4
For (b): Sodium chloride dissociates into ions not present in the equilibrium. There is no common ion, so pH stays the same for AP purposes.

Exam tip:

Always write the balanced equilibrium reaction in your justification; AP graders require explicit reference to shift direction for full credit.

5. AP Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding with these AP-style questions:

Which of the following changes will result in a decrease in the percent ionization of 0.20 M ammonia ( $N H_{3}$ , a weak base) due exclusively to the common ion effect?
- Adding pure water to dilute the solution
- Adding 0.10 mol of solid ammonium nitrate ( $N H_{4} N O_{3}$ ) to the solution
- Adding 0.10 mol of solid sodium chloride (NaCl) to the solution
- Adding 0.10 mol of solid potassium nitrate ( $K N O_{3}$ ) to the solution
Reveal answer
1 —
Correct. Ammonium nitrate provides the common ion $N H_{4}^{+}$ , shifting equilibrium left and reducing percent ionization. Dilution increases percent ionization, and the other options have no common ions.
A solution contains 0.30 M propanoic acid ( $K_{a} = 1.3 \times 1 0^{- 5}$ ) and 0.20 M calcium propanoate. What is the pH of the solution?
- 4.79
- 4.89
- 5.01
- 5.12
Reveal answer
2 —
Correct. Calcium propanoate dissociates to give 0.40 M propanoate. $p H = 4.89 + lo g (0.40/0.30) = 5.01$ .

6. Common Pitfalls

Wrong move:

Calculating the common ion concentration by adding the full contribution of the dissolved weak electrolyte/sparingly soluble salt to the added common ion concentration

Why:

Students forget that dissociation/dissolution is already suppressed, so the contribution from the weak electrolyte is negligible

Correct move:

Always apply the small x approximation first; only add the contribution if the approximation fails (almost never happens on AP problems)

Wrong move:

Predicting that adding a common conjugate base to a weak acid decreases pH

Why:

Students confuse lower $[H^{+}]$ with lower pH, or mix up the direction of equilibrium shift

Correct move:

Always write the equilibrium before predicting: adding product ion shifts left, so $[H^{+}]$ decreases, and pH increases

Wrong move:

For $C a F_{2}$ in NaF solution, writing $K_{s p} = (s) (0.10)$ forgetting the stoichiometric coefficient for $F^{-}$

Why:

Students memorize the 1:1 salt formula and apply it to all salts regardless of stoichiometry

Correct move:

Always balance the dissolution reaction and map equilibrium concentrations to solubility before plugging into $K_{s p}$

Wrong move:

Claims adding any salt to a weak electrolyte solution triggers the common ion effect

Why:

Students assume all salts contain a common ion, but only salts with one ion matching the equilibrium produce the effect

Correct move:

Check both ions of the added salt against the equilibrium to confirm one is common before invoking the effect

Wrong move:

Swapping the concentrations of weak acid and conjugate base in the Henderson-Hasselbalch equation

Why:

Students misremember the order of terms in the formula

Correct move:

Derive $[H^{+}]$ directly from the $K_{a}$ expression if you cannot remember the order of terms

7. Quick Reference Cheatsheet

Category	Formula	Notes
Weak acid + common conjugate base	$p H = p K_{a} + lo g (\frac{[ A ^{-} ] _{0}}{[ H A ] _{0}})$	Valid when 5% approximation holds, almost always true for AP problems
Weak base + common conjugate acid	$pO H = p K_{b} + lo g (\frac{[ B H ^{+} ] _{0}}{[ B ] _{0}})$	Convert pOH to pH via $p H = 14 - pO H$ at 25°C
$M_{x} X_{y}$ with common M ion	$K_{s p} = ([M]_{0})^{x} (y s)^{y}$	s = molar solubility; approximation $x s << [M]_{0}$ applies
$M_{x} X_{y}$ with common X ion	$K_{s p} = (x s)^{x} ([X]_{0})^{y}$	s = molar solubility; approximation $y s << [X]_{0}$ applies
Equilibrium shift rule	Adding common product ion → shift toward reactants	Applies to both acid-base ionization and dissolution
Percent ionization change	Adding common ion → percent ionization decreases	Always true for weak electrolytes

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Common ion pH change prediction
2022 · FRQ
Solubility calculation with common ion

Going deeper

Official SyllabusAP Chemistry Course and Exam DescriptionUnit 7 Equilibrium section

What's Next

The common ion effect is the foundational prerequisite for buffer solutions, the next major topic in Unit 7 Equilibrium. Buffers are literally common ion systems (weak acid + common conjugate ion, or weak base + common conjugate ion) designed to resist pH change, so without understanding how common ions suppress ionization and shift equilibrium, you will not be able to calculate buffer pH or predict buffer capacity. Beyond buffers, the common ion effect is also a core concept for understanding solubility equilibria, selective precipitation, and titration pH curves, all of which are tested heavily on the AP Chemistry exam.